Leetcode: palindrome partitioning II [0, 132]

[Question]

Given a string S, partition S such that every substring of the partition is a palindrome.

Return the minimum cuts needed for a palindrome partitioning of S.

For example, given S ="aab",
Return1Since the palindrome partitioning["aa","b"]Cocould be produced using 1 cut.

Question]

Given a string S, the minimum number of splits is returned. Yes, all the substrings are input.

[Idea]

DP, and determine the minimum number of Split times of S [1... n] in sequence
Assume that mincuts [I] is used to represent the minimum number of times the string ending with I is split.
So how to determine mincuts [I]. We know mincuts [J] J =,..., I-1
For each position J before I, if s [J + 1,..., I] is the input string, the minimum number of times J is cut at the cutting point is mincuts [J] + 1;
Therefore, we need to take a position from all J so that the position with the smallest number of cut times at the I position is used as the cutting point, mincuts [I] = min {mincuts [J] + 1} s [J + 1 ,.., i] is input, j = 0, 1... i-1

[Code]

Class solution {public: int mincut (string s) {int Len = S. length (); If (LEN = 0 | Len = 1) return 0; // construct the echo discriminant matrix vector <bool> ispal (Len, vector <bool> (Len, false); // initialize ispal [I] [I] for (INT I = 0; I <Len; I ++) ispal [I] [I] = true; // initialize ispal [I] [I + 1] For (INT I = 0; I <len-1; I ++) if (s [I] = s [I + 1]) ispal [I] [I + 1] = true; // initialize other I <j substrings for (INT I = len-3; I> = 0; I --) for (Int J = I + 2; j <Len; j ++) ispal [I] [J] = (s [I] = s [J]) & Ispal [I + 1] [J-1]; // returns the smallest cut vector <int> mincuts (Len, 0 ); // record the minimum cut score of the substring ending with I for (INT I = 0; I <Len; I ++) {int mincut = int_max; for (Int J =-1; j <I; j ++) {If (ispal [J + 1] [I]) {If (j =-1) mincut = 0; else if (mincuts [J] + 1 <mincut) mincut = mincuts [J] + 1 ;}} if (mincut! = Int_max) mincuts [I] = mincut;} return mincuts [len-1];};